KNOWLEDGE TO GET FROM TODAY S CLASS MEETING

Transcription

1 KNOWLEDGE TO GET FROM TODAY S CLASS MEETING Class Meeting #6, Monday, February 1 st, ) GRAVITY: finish up from Fri, Jan 29 th (pages , 123) 2) Isaac Newton s LAWS of MOTION (briefly) (pages ) 3) Distances of interest (pages 2, 4, 15, 70, A-15) 4) Planet Orbit characteristics (shapes) (pgs 65-67) 5) Kepler s 3 Laws of Planetary Motion (these are important!!) (pgs 66-70)

2 Your Lab Sections Wednesday & Thursday in Room 232 in Walden Hall will conduct the DENSITY Lab Read through the lab BEFORE your section meets!! ANY QUESTIONS? You should have a printed copy of the DENSITY Lab with you when you arrive at your Lab Section meeting The PDF of the Lab is available within the LAB-MANUAL folder at the class website: Spring2016/LAB-MANUAL/LAB02-Density-Feb03-04.pdf

3 QUIZ #2 WILL OCCUR: this coming Friday, February 5 th (the final ~20 minutes of class) Topics: i) Kepler s 3 Laws of Planet Motion ii) initial telescopic observations which supported the Sun-centered theory of Solar System structure iii) Eratosthenes determination of Earth s size iv) concepts of Mass, Volume, Density, & Gravity v) the 8 planets in our solar system and their order of increasing Orbital Semi-Major Axis value: Table E1, pg A-15 listed on the next slide are text sections and questions for quiz study

4 Kepler s Laws: text pgs 66-68, 70, 73,79 Text questions: Chapt 3 #s 9, 20, 25, 26, 27 Initial Observations that supported Copernicus idea: text pg Text questions: Chapt 3 #s 723, 24, 29 Eratosthenes: text pg 63 Text questions: Chap 3 #s 45, 46 (if you can do the math, great, but the idea is what s important) Mass, Volume, Density: text pgs ; (in Chap 5) Text questions: Chap 4 #s 4, 11, 15, 26 ; Chap 5 #s 6, 7, 30 Gravity (including comparing mass to weight: text pages Text questions: Chap 4, #s 9, 18, 19, 32, 33, 42

5 You should by the end of this week have read through CHAPTER 4 in the text.. and pages in Chapt 5 (with special emphasis upon the topics we have discussed here in class PLUS Eratosthenes method for estimating Earth s circumference, and Galileo s Copernicussupporting telescope observations, both in Chapter 2)

6 RESULTS OF QUIZ #1 Average = 16.9/20 (84.5%) Median = 17

7 LET S RETURN TO OUR TOPIC OF GRAVITY from our previous class but first ANY QUESTIONS????

8 1) The Gravitational attraction (force) between YOU (70 kg) and the Earth is: F Grav = G Mass Earth Mass You / (Earth s radius) 2 = 6.67 x N m 2 /kg 2 x 6 x kg x 70 kg (6,378,000 meters) 2 = Newtons (equivalent to 154 pounds)

9 When we calculate the gravitational force between the Earth and another object, we assume that all of Earth s mass is concentrated at its center:

10 It is sometimes easier to think in terms of: Gravitational Acceleration = F Grav / Mass You if one mass is much, much larger than the other Earth s downward acceleration at its surface: Newtons / 70 kg = 9.8 (m/s)/s Earth s Surface Gravitational Acceleration = 9.8 (m/s)/s Which indicates that if you, or anything else, is dropped here at the Earth s surface, its FALLING SPEED will increase by 9.8 meters per second (~20 mph) during each second that you fall

11 Acceleration m/s (9.8 m/s)/s Time = 0 seconds, Speed = 0 m/s (time to fall 1 meter = 0.45 seconds) Time = 1 second, Speed=9.8 m/s Distance fallen = 4.9 meters ( (= ~16.5 feet) 44.1 meters Time = 2 seconds; Speed=19.6 m/s Distance fallen=19.6 meters (= ~66 feet) Time = 3 seconds; Speed=29.4 m/s Distance fallen=44.1 meters (= ~150 feet)

12 What is the Gravitational Acceleration rate between two people seated side-by-side in this classroom? F Grav = G Mass Neighbor Mass You / (1 meter) 2 = 6.67 x N m 2 /kg 2 x 70 kg x 70 kg (1) 2 = Newtons (=3.27 x 10-7 Newtons) Gravitational Acceleration between you two: 3.27x10-7 Newtons / 70 kg = 5 x 10-9 (m/s)/s (equal to ~ miles per hour per second)

13 SO, due to gravity: i) an object dropped from 1 meter above the ground will require 0.45 seconds to complete its fall BUT ii) Two people 1-meter apart from each other and far out in space away from any large mass (source of gravity) would require approximately 20,000 seconds (approximately 5 hrs 33 minutes) before their centers collide For Gravity Acceleration conditions: DISTANCE = ½ x Acceleration x TIME 2

14 So, how can these concepts be used to understand the Planets (or moons or stars)? 1) you want to know what material(s) some planet in the solar system is composed of; you can see it, and know its size (radius) (from which you can determine its VOLUME), but don t know much more. 2) if a moon orbits the object, or a spacecraft passes by, you can determine the gravitational pull of the planet upon the moon or upon the spacecraft; a measurement of the GRAVITY can then permit you to calculate the MASS of the planet

15 Consider a planet in the Solar System; a spacecraft is going to pass by the planet The spacecraft will feel the planet s gravitational attraction ( acceleration induced upon the craft) The magnitude of the gravitational attraction felt, and the known distance between the planet and the spacecraft, allow us to determine the planet s MASS A moon orbiting a planet can also provide information from which the planet s mass, and density can be determined

16 3) If you know the VOLUME of the planet and its MASS, you can determine the DENSITY of the planet: DENSITY = MASS / VOLUME 4) Planets are not pure substances, rather they are a mixture of many materials, but a planet s DENSITY can tell you if it is rocky:metallic (like Earth) or gaseous hydrogen & helium (like Jupiter), WITHOUT EVER HAVING TO GO TO THAT PLANET!!

17 This technique allowed astronomers to learn Jupiter s density, ~1.3 grams per cubic centimeter, well before we ever sent a spacecraft to Jupiter.. The telescopically observed apparent size of Jupiter and our knowledge of its distance from the Sun, PLUS the orbit characteristics of Jupiter s four large moons, permitted determination of Jupiter s volume, its gravitation attraction upon its moons, and thus Jupiter s mass, and.. DENSITY = MASS / VOLUME

18 ANY QUESTIONS ABOUT GRAVITY? ORBIT MOTIONS (planets around the Sun, the Moon around Earth, etc.) are controlled by the Gravity Force between the central object and the orbiting object Let s start to take a look at this control. with Isaac Newton s 3 Laws of Motion

19 Isaac NEWTON S 3 Laws of Motion (pg 115) (these help us understand orbit motions.) 1 st Law: an object continues at the same speed (which can be zero) and in the same straight direction unless acted upon by an outside force 2 nd Law: the acceleration (change in speed OR direction) an object experiences is directly related to the magnitude of the force applied upon the object & the mass of the object 3 rd Law: each action (force) occurs in the presence of an equal magnitude but oppositedirection action (force) [ action:reaction ].. we will in a few minutes discuss how gravity and Newton s laws of motion control orbit motions

20 Increasing Length OK, let s recall DISTANCE UNITS for a moment METRIC ENGLISH Units of length 1 centimeter (10 millimeters; inch) 1 inch (2.54 centimeters) 1 foot(12 inches; cm) 1 yard (3 feet; cm) 1 meter (100 cm; inches) 1 kilometer (1000 meters; miles) 1 mile (1.61 kilometers) 5 kilometers (3.1 miles)

21 ASTRONOMICAL DISTANCES Within our Solar System, the Standard Distance is the average Sun-to-Earth distance: ~93,000,000 miles = ~9.3 x 10 7 miles [ How many kilometers is this equal to? ] 9.3 x 10 7 miles x 1.61 km per mile ~1.5 x 10 8 km (which is 391 times the Earth-to-Moon distance) This distance (150,000,000 km or 93,000,000 miles) is called: ONE ASTRONOMICAL UNIT (1 AU)!!! 1 AU Earth

22 In Our SOLAR SYSTEM, Planets Average Distance from the Sun range from: Closest to the Sun MERCURY: AU (= 6 x 10 7 km = 37 million miles) to Farthest from the Sun PLUTO: AU (= 6 x 10 9 km = 3.7 billion miles) (and remember, Earth is 1 AU from the Sun)

23 While AUs are a very nice length scale to employ when talking about distances within our Solar System (better than meters, or feet, or leagues) AU s are not very useful when discussing distances BETWEEN stars within our Milky Way Galaxy FOR EXAMPLE: The Sun s nearest stellar neighbor, a star named ALPHA CENTAURI, is located: ~250,000 AU (= 2.5 x 10 5 AU) from the Sun What other Astronomical unit(s) are you familiar with?

24 PARSEC= 3.1 x kilometers (=210,000 AU) LIGHT YEAR= 9.5 x km (= 63,000 AU) (so. 1 PARSEC = 3.3 LIGHT YEARS) What does a LIGHT YEAR really mean? ( is it a time duration, or a distance, or.? ) A TIME = DISTANCE EXAMPLE: How far is it from Las Cruces to Deming? 97 kilometers (60 miles) But, what if I said, It is one hour to Deming. How would you interpret this?

25 Travelling at freeway speed, on I-10, a time interval of one hour will get me to Deming So, a time interval at a known speed = a distance This is the idea behind the distance unit of a LIGHT YEAR ONE LIGHT YEAR = the distance light travels in one year = SPEED OF LIGHT times a time interval of 1 YEAR = 3 x 10 8 meters per second x days per year x 86,400 seconds per day = 9.5 x meters (= 9.5 x km = 5.9 x miles).. and one AU = 1.5 x 108 km or 93,000,000 miles, so..

26 ONE LIGHT YEAR = ~ 63,000 AU Speed of light = 186,000 miles per second = 3 x 10 8 meters per second How far apart are the Sun and Earth? 1 AU = 1.5 x 10 8 kilometers = 93,000,000 miles How about their distance in LIGHT TIME? ~8 minutes! So, sunlight striking the Earth s surface here in Las Cruces right now left the Sun 8 minutes ago.. and, the nearest star beyond the Sun is 4 Light Years away!!

27 Since the Speed of Light is the fastest speed physics allows, if the Sun was to shut off right now, we would not know about it for ~8 minutes Mercury (0.4 AU) is 3.2 light minutes from the Sun, Pluto (40 AU) 320 minutes or 5.33 hours from the Sun ANY QUESTIONS ABOUT GRAVITY, or DISTANCES?

28 NOW, on to Johannes Kepler s determination of three (3) characterizations of planetary motions around the Sun THESE ARE VERY IMPORTANT CONCEPTS! Kepler developed these concepts BEFORE the mathematical understanding of gravity was appreciated Kepler developed these Laws in the late 1500 s and early 1600 s

29 JOHANNES KEPLER s Early 1600 s 3 LAWS OF PLANETARY MOTION Kepler developed his Laws by analyzing many years of eyeball observations of planets changing positions relative to the background pattern of stars (Mars motions were especially useful!!) Kepler was a believer in Copernicus idea that the Earth traveled (orbited) around the Sun (and so did the other known 5 planets: Mercury, Venus, Mars, Jupiter, Saturn) Copernicus Sun-centered theory arose in the 1540 s

30 The Sun-centered structure of the Solar System, and measured angles of planet positions, enabled Kepler to calculate the distance between a planet and the Sun, relative to the Earth-Sun distance, equal to 1 AU Mars Mars Orbit 1.52 AU Earth 1 AU Measure this angle calculate that length Earth s Orbit The angle at the Sun is a 90-degree angle

31 When Mars and Earth are at different orbit locations, another measure of Mars distance from the Sun can be determined (and another, and then another, ): Mars orbit This is a 90-degree angle Earth s orbit 1 AU 1.48 AU Earth Mars Measure this angle calculate that length

32 1) Kepler s 1 st Law of Planetary Motion: The shape of a planet s orbit around the Sun is an ELLIPSE (a circle is one type of an ellipse, but no planet orbit in our Solar System is a perfect circle) C Orbit A: circular (eccentricity = 0) Orbit B: ellipse (eccentricity = ~0.5) Orbit C: ellipse (eccentricity = ~0.9) B A All 3 of these planets have the same Orbital Semi-Major Axis, which is the measure of the AVERAGE distance between the planet and the Sun

33 An ELLIPSE is defined by its ECCENTRICITY ( non-circularity ) and its SEMI-MAJOR AXIS (SMA) length Eccentricity values range from ZERO (for a circle) to (for a really stretched out ellipse) [Figure 3.18 in text] Major Axis The Semi-Major axis (SMA) length is equal to ½ of the Major Axis Length and for a planet orbiting the Sun, it s SMA is that planet s AVERAGE distance from the Sun F2 F1

34 In the process of determining his 1 st Law, Kepler also determined that for a SINGLE PLANET travelling along an elliptical (non-circular) orbit, that single planet traveled fastest when nearest to the Sun, and slowest when farthest from the Sun. This understanding is now: Figure 3.20 in your text KEPLER S 2 ND LAW OF PLANETARY MOTION: while travelling on their elliptical orbits, an individual planet travels fastest when nearest to the Sun, and slowest when farthest from the Sun (this is also known as the Equal Area in Equal Time Law) [demonstrated on-line with your text, Figure 3.18]

35 Kepler s 2 nd Law A B See Figs 3.16 and 3.18 in your text C Cp Bp Location of slowest speed in orbit Orbit A: circular (orbital speed is constant) Orbit B: ellipse (planet travels fastest at point Bp) Orbit C: ellipse (planet travels fastest at point Cp) We can animate Kepler s 2 nd law at:

36 But.. KEPLER S 2 ND LAW DOES NOT tell us anything directly about how long a given planet takes to complete one orbit. For ORBIT PERIOD information we must use KEPLER S 3 RD LAW OF PLANET MOTION

37 Using the available observations, Kepler was also able to determine that: planets farther from the Sun on average (planets with larger Orbital Semi-Major Axes) require more time to complete one trip (one orbit) around the Sun than do planets nearer to the Sun (planets with smaller OSA s). Using his knowledge of the orbital semi-major axes of the planets (relative to Earth s = 1 AU) and also their orbital periods, Kepler worked to find a mathematical relationship between Orbital Semi-Major Axis and Orbital Period let s consider an incorrect idea first..

38 IF ORBITAL PERIOD depended directly upon ORBITAL SEMI-MAJOR AXIS, ( twice as far = twice as long ) then: ORBITAL PERIOD OSA Estimated Earth 1 AU 1 Earth Year Mercury 0.4 AU 0.4 EY(146d) Venus 0.7 AU 0.7 EY (255d) Mars 1.52 AU 1.52 EY Jupiter 5.2 AU 5.2 EY Saturn 9.5 AU 9.5 EY These results would arise IF all of the planets traveled at the same speed BUT actually the larger a planet s OSA, the smaller its average orbit speed

39 The OBSERVED orbit periods for planets indicate that planets farther from the Sun travel more slowly (on average) than does Earth, and planets closer to the Sun (on average) travel faster than Earth ORBITAL PERIOD OSA Estimated Observed Earth 1 AU 1 Earth Year 1 Earth Year Mercury 0.4 AU 0.4 EY(146d) 0.25 EY (88d) Venus 0.7 AU 0.7 EY (255d) 0.58 EY (224d) Mars 1.52 AU 1.52 EY 1.88 EY Jupiter 5.2 AU 5.2 EY ~12 EY Saturn 9.5 AU 9.5 EY ~30 EY These results indicate that planets do not all travel at the same speed, rather planets farther from the Sun than Earth travel more slowly than does Earth

40 This Relationship between distance from the Sun and orbital speed, and thus Orbit Period, is quantified in Kepler s 3 rd Law, which relates: i) how far a planet is from the Sun on average (its Orbital Semi-major Axis distance, OSA) VS. ii) the length of time for that planet to complete one complete orbit around the Sun, the planet s ORBIT PERIOD:

41 KEPLER s 3 rd LAW of Planetary Motion: (Orbit Period) 2 = OSA 3 Period x Period = OSA x OSA x OSA with PERIOD in units of Earth years and OSA in units of Astronomical Units GRAVITY is responsible for Kepler s Laws

42 For Example: Orbital Period vs. Orbital Semimajor Axis P 2 = OSA 3 which leads to P = \ OSA 3 ) OSA Period (Earth years) Mercury: 0.39 AU Venus: 0.72 AU Earth: 1.0 AU Mars: 1.52 AU Jupiter: 5.20 AU Saturn: 9.54 AU Uranus: 19.2 AU Neptune: 30.1 AU Pluto: 40 AU

43 For Example: Orbital Period vs. Orbital Semimajor Axis P 2 = OSA 3 which leads to P = \ OSA 3 ) OSA Period (Earth years) Mercury: 0.39 AU 0.24 Venus: 0.72 AU 0.62 Earth: 1.0 AU 1.0 Mars: 1.52 AU 1.88 Jupiter: 5.20 AU Saturn: 9.54 AU Uranus: 19.2 AU 84.1 Neptune: 30.1 AU 165 Pluto: 40 AU 253 Notice: we DO NOT use Eccentricity for Period!!!!

44 NOTE: Kepler s 3 rd Law DOES NOT CARE about the eccentricity of the orbit in question, it CARES ONLY about the Orbital Semi-Major Axis (average distance to the Sun during an orbit) A C B All 3 of these planets have the same Orbital Semi- Major Axis length, which is the measure of the average distance between the planet and the Sun Orbit A: (eccentricity = 0; OSA=1AU; Period=1EY) Orbit B: (eccentricity=0.5; OSA=1AU; Period=1EY) Orbit C: (eccentricity=0.9; OSA=1AU; Period=1EY)

45 TABLE of PLANET INFORMATION

46 KEPLER S 3 LAWS OF PLANETARY MOTION 1) Planet orbits are elliptical in shape (includes circular) 2) When traveling along its elliptical orbit, a planet travels at its fastest speed when it is closest to the Sun (and slowest when farthest from the Sun) 3) P 2 = OSA 3 with P in units of Earth Years and OSA in units of Astronomical Units 1 Astronomical Unit = 1 AU 1 AU = Earth s orbital semi-major axis Let s work through an exercise to illustrate these concepts

47 Now, back to Isaac NEWTON S 3 Laws of Motion (these help us understand orbit motions.) 1 st Law: an object continues at the same speed (which can be zero) and in the same direction unless acted upon by an outside force 2 nd Law: the acceleration (change in speed OR direction) an object experiences is directly related to the magnitude of the force applied upon the object and the mass of the object 3 rd Law: each action (force) occurs in the presence of an equal magnitude but opposite direction action (force) [ action:reaction ]

48 Isaac Newton s 3 Laws of Motion help us understand Kepler s Laws NEWTON S FIRST LAW An object moves at a constant velocity (speed and direction) if there is no net force acting upon it. The constant velocity can be zero speed, so an object held in your hand has a velocity of zero (relative to the ground) because the upward force provided by your hand is negating the dominating downward force of Earth s gravity (no NET force is acting upon the object)

49 NEWTON S SECOND LAW Force = Mass times Acceleration A potentially easier way to think of this is: Acceleration = Force / Mass If you drop the object in your hand (remove the upward force you are providing), the object will accelerate downward, due to the force of Earth s dominating gravitational force attraction, and the object s acceleration will be 9.8 (m/s)/s, which we discussed during our previous class meeting

50 NEWTON S THIRD LAW For any force, there is always an equal and opposite reaction force This law is not as easily demonstrated as are Newton s first two laws the Earth is exerting a gravitational force upon you AND you are exerting a gravitational force upon the Earth.. If this is true (and it is), then WHY when you step off of a tall building do you appear to fall to the Earth s surface, rather than having the Earth s surface rise to you? this discussion is continued on next slide

51 As Newton s 2 nd Law tells us: Acceleration = Force / Mass As we calculated previously, the FORCE between 70 kg you and the Earth is: 688 Newtons and the gravitational acceleration you feel due to the Earth is: 688 N / 70 kg = 9.8 (m/s)/s (~20 mph/second) BUT the acceleration the Earth feels from you is only: 688 N / (6 x kg) = 1.1 x (m/s)/s (which is equal to two tenth billionths of one trillionth of a mile per hour per second!!!).. this is Newton s 3 rd Law!!

52 The launching of a rocket is an example of Newton s 3 rd Law of Motion in action: THE ROCKET S MASS MOVING UPWARD Hot gas molecules accelerated downward by the burning of the fuel result in a reaction that is the upward acceleration of the rocket HOT GAS ATOMS and MOLECULES moving downward

53 So, how can NETWON S 3 Laws and Gravity help us understand Kepler s Laws? Newton s 1 st and 2 nd Laws tell us that a planet would travel along a straight line path (constant speed and direction) IF no external (unbalanced) force was acting upon the planet

54 BUT We know planets travel along curved ellipseshaped paths (orbits), indicating that some force must be acting upon the planet. the force is GRAVITY between planet and Sun GRAVITY

55 We can think of orbital motion as a balance (Newton s 3 rd Law of Motion) between the: GRAVITY FORCE vs CENTRIPETAL FORCE F G = F C F G F C x Speed 2 = M P Orbit Radius

56 NEWTON S MODIFIED VERSION OF KEPLER S 3 rd LAW Kepler s 3 rd Law is: P 2 = OSA 3 It WORKS only for an object orbiting our Sun, OR for an object orbiting around another star which possesses the exact same mass as (same number of protons and neutrons as) the Sun Isaac Newton determined, after he conceived of the physical understanding of GRAVITY (~100 years after Kepler determined his 3 rd Law), that Kepler had made a slight error

57 Newton determined that the true form of Kepler s 3 rd Law should be: (Mass Sun + Mass Planet ) x P 2 = OSA 3 Mass Sun..because the Sun and the planet exert a gravitational attraction upon EACH OTHER!.. it is NOT just one way!! Why did KEPLER miss this mass issue? Mass Sun + Mass Jupiter = Not much different from a value of 1.0!! Mass Sun.. so the mass of the planet(s) is inconsequential, and so is its pull upon the Sun!!

58 One item of information this MODIFIED version of KEPLER S 3 RD LAW provides is the following: If the Sun was more massive than it actually is, than the Earth s orbital period would be < 1 Earth year, and if the Sun was less massive than it actually is, than the Earth s orbital period would be > 1 Earth Year We can also use this MODIFIED version of KEPLER S 3 RD LAW to calculate some orbital information, and more.

59 [(M Star + M planet ) / M Sun ] x P 2 = OSA 3 Knowing that the Earth s mass is << the Sun s Mass: (so we can ignore M planet above) If the Sun s mass was twice its actual value: M Sun x P Earth = OSA Earth M Sun 1 2 x P Earth 2 = OSA Earth 3 P 2 = 1/2 P= 1/2 P Earth = 1/1.414 = EY If M Sun was only 1/2 of the Sun s mass, P = EY

60 As it turns out, the modified form of Kepler s 3 rd Law is a very powerful tool that can be applied to any two objects orbiting each other (not just planets around stars). GENERAL FORM OF MODIFIED 3 rd LAW (M central + M orbiter ) x P 2 =OSA orbiter 3 M Sun 1 AU We can use this equation to calculate the Moon s orbital period around the Earth (assuming we know the Moon s OSA) (applicable when M central >> M orbiter )

61 What is the Moon s orbital period around Earth? (M Earth + M Moon ) x P 2 = OSA Moon 3 M Sun M Earth = 6 x kg; M Moon = 7.35 x kg 1 AU M Sun = 2 x kg; OSA Moon = 384,00 km= AU [(6 x x ) /(2 x )] x P 2 = ( ) 3 P 2 = Earth years P Moon = Earth years = 27.3 days Compare this to information on page A16 (Appendix E) in your text..

62 KEPLER S 3 LAWS OF PLANETARY MOTION 1) Planet orbits are elliptical in shape 2) When traveling along its elliptical orbit, a planet travels at its fastest speed when it is closest to the Sun (and slowest when farthest from the Sun) 3) P 2 = OSA 3 with P in units of Earth Years and OSA in units of Astronomical Units 1 Astronomical Unit = 1 AU 1 AU = Earth s orbital semi-major axis GRAVITY is the force that dictates these Laws!

63 Kepler s 3 rd Law and Gravity The following steps illustrate how Kepler s 3 rd Law of Planetary Motion is related to Newton s formulation of Gravity; both GRAVITATIONAL force and CENTRIPETAL force (related to curved, rather than straight, motion): 1) Sun s Gravitational force: F grav = G M Sun M planet / [ OSA(in meters) 2 ] 2) Planet s Centripetal (think centrifugal ) force: F cent = M planet * Speed planet 2 / OSA (in meters)

64 We can think of orbital motion as a balance between the: GRAVITY FORCE vs CENTRIPETAL FORCE F G = F C F G F C = M P Orbit Radius Speed 2 Speed = Distance around orbit circle Orbit Period

65 3) For balance: F grav = F cent (G M Sun M planet / OSA 2 ) = M planet * Speed planet 2 / OSA 4) Speed = distance around orbit divided by time Speed = 2 * π * OSA / (Orbit Period) 5) So: Gravitational Force = Centripetal Force M Planet M Planet = G M Sun / OSA 2 = (4 π 2 OSA 2 / Orbit Period 2 ) / OSA 6) Rearranging: (G M Sun / 4 π 2 ) x (Orbit Period) 2 = OSA 3 Constant x P 2 = OSA 3

66 7) (G M Sun / 4 π 2 ) x (Orbit Period) 2 = OSA 3 Up to this point, we have been using units of kilograms, meters, and seconds; if we stick in the correct values for G and M Sun and π, and convert Orbit Period to Earth years and OSA to AU, we get: P 2 = OSA 3 with Period in Earth years, OSA in AU SO, KEPLER S LAWS WORK IN EXPLAINING PLANET ORBIT MOTIONS BECAUSE OF THE NATURE OF THE GRAVITATIONAL FORCE